试求Y=2cos^2x 5sinx-4点最值

楼上的少写了“-”和“dx”吧dy=2cos(x+1)•[-sin(x+1)]dx=-sin2(x+1)dx

y=sinxcosx-cos^2x=1/2sin2x-1/2(1+cos2x)=1/2(sin2x-cos2x-1)=1/2[√2*sin(2x-派/4)-1]=√2/2*sin(2x-派/4)-1/

cosα+cosβ=cosα+cos(120-α)利用和差化积公式:cosα+cos(120-α)=2cos[(120-α+α)/2]*cos[(120-α-α)/2]=cos(60-α)=1/2(x

y=(cosx+2)/(sinx-1)ysinx-y=cosx+2ysinx-cosx=y+2√(y²+1)sin(x-t)=y+2,t=arctan(1/y)sin(x-t)=(y+2)/

就是简单的复合函数求导问题嘛.1.y'=[1/cos(10+2x）]*[-sin(10+2x)]*2=[-2sin(10+2x)]/cos(10+2x)2.y'=[1/cos(3+x²)]*

COS（X+Y）COS（X-Y）=(COSX*COSY-SINX*SINY)(COSX*COSY+SINX*SINY)=(COSX*COSY)^2-(SINX*SINY)^2=COS^2X(1-SIN

5/9cos(x-y)=cosx*cosy+sinx*sinysin(x)-sin(y)=-(2/3),两边平方得到sin^2x-2sinxsiny+sin^2y=4/9cos(x)-cos(y)=(

y=cos^2x+cosx^2y'=2cosx(-sinx)+(-sinx^2)*2x=-2sinxcosx-2xsinx^2=-sin2x-2xsinx^2

y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos

y=cos^2x+sinx=1-2(sinx)^2+sinx=-2(sinx-1/4)^2+9/8因为|x|

y=[cosx-1-1]/(cosx-1)=1-1/(cosx-1)=1-1/(1-2sin^2(x/2)-1)=1+1/(2*sin^2(x/2))故其周期是T=2π

-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k

Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2

先对cos求导=-sinx^2再对x^2求导=2x所以y'=-2x*cosx^2

2sin^2x+cos^2y=1cos^2y=1-2sin^2x≥0∴0≤1-2sin^2x≤1∴0≤sin^2x≤1/2∴sin^2x+cos^2y=sin^2x+1-2sin^2x=1-sin^2

y=(sinx+cos)^2+2cos^2x=1+2sinxcosx+cos2x-1=sin2x+cos2x=√2sin（2x+π/4)

cos²A+cos²C=(cos2A+cos2C+2)/2=[2cos(A+C)cos(A-C)+2]/2=cos(A+C)cos(A-C)+1=1-cos(A-C)/2上式要有最

y=cosB+sin(B/2)=1-2sin@（B/2）+sin（B/2)=(0,9/8)@表示平方求范围用（）@的方法注意B是三角形中一个角的特殊情况所以B/2的范围是（0,π/2)

cos2x=cos方x-sin方x=2cos方x-1cos方2x=(cos4x+1)/2T=2兀/4=兀/2